The probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is [tex]\frac{5}{12}[/tex].
Solution:
Total number of outcomes N(S) = 36
Let A be the sum that is a multiple of 6 and
B be the sum that is a multiple of 4.
Sum that is a multiple of 6 = (1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6)
N(A) = 6
Sum that is a multiple of 4 = (1, 3), (2, 2), (2, 6), (3, 1), (3, 5),
(4, 4), (5, 3), (6, 2), (6, 6)
N(B) = 9
[tex]$P(A)=\frac{N(A)}{N(S)}[/tex]
[tex]$P(A)=\frac{6}{36}=\frac{1}{6}[/tex]
[tex]$P(B)=\frac{N(B)}{N(S)}[/tex]
[tex]$P(B)=\frac{9}{36}=\frac{1}{4}[/tex]
Probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4:
[tex]P(A \cup B)=P(A)+P(B)[/tex]
[tex]$P(A \cup B)=\frac{1}{6} +\frac{1}{4}[/tex]
[tex]$=\frac{2+3}{12}[/tex] (Make the denominator same)
[tex]$=\frac{5}{12}[/tex]
Hence the probability that the outcome is a sum that is a multiple of 6 or a sum that is a multiple of 4 is [tex]\frac{5}{12}[/tex].
